I don't think that was so much a "hint" as an "Apply x as $5+5\sin{t}$".
–
NeilApr 24 '12 at 9:44

2

If you substitute $x=5+5sin(t)$, then you must note that $dx=5cos(t)dt$, which you forgot from your calculations. Right now you're integrating $t$ respect to the variable $x$.
–
Thomas E.Apr 24 '12 at 9:46

Ok, So I have ... $$5\int\sqrt{1-(\frac{t}{5})^2}=5\sin^{-1}{\frac{x-5}{5}}$$ Is it right? I also tried letting $t=\frac{x-5}{5}, dx=5 dt$ $$5 \int \sqrt{1-t^2}\cdot 5 dt=25 \sin^{-1}{\frac{x-5}{5}}$$ They appear different? I must me making some very stupid mistakes again? $$$$
–
Jiew MengApr 24 '12 at 10:27